Hydrogen gas is a very environmentally friendly fuel source. One factor that is important in determining whether it is used as a fuel source is the cost of production. Use the two equations below to evaluate which reaction requires more energy to produce 1 mol of H2(g). Reaction (1) as written produces H2(g) from methane and is 80.5% efficient. The production of H2(g) by electrolysis of water, as shown in reaction (2), is 49.1% efficient.

Question

Hydrogen gas is a very environmentally friendly fuel source. One factor that is important in determining whether it is used as a fuel source is the cost of production. Use the two equations below to evaluate which reaction requires more energy to produce 1 mol of H2(g). Reaction (1) as written produces H2(g) from methane and is 80.5% efficient. The production of H2(g) by electrolysis of water, as shown in reaction (2), is 49.1% efficient.

(1) CH4 (g) + H2 O(l) → 3H2 (g) + CO(g) ∆H° r = +205.9 kJ
(2) H2O(l) → H2 (g) + _1 2 O2 (g) ∆H° r = +285.8 kJ

Answer 1

So 205.9 kJ is needed to produce 3; 205.9/3 kJ will be needed to produce 1 mol and since that rxn is only 80.5% efficient then 205.9/3/0.805 will be needed to produce 1 mol at that level of efficiency. Do the same for the H2O electrolysis and compare.

Answer 2

To determine which reaction requires more energy to produce 1 mole of H2(g), we need to calculate the energy input for each reaction and compare them.

Let's start with reaction (1): CH4(g) + H2O(l) → 3H2(g) + CO(g)

From the given equation, the reaction enthalpy (∆H°r) is +205.9 kJ. This means that 205.9 kJ of energy is released for every mole of CH4 and H2O reacted to produce 3 moles of H2. However, we also need to consider that this reaction is 80.5% efficient. Efficiency refers to the ratio of actual output (useful energy obtained) to the theoretical maximum output (energy released in the reaction).

Therefore, the energy input for reaction (1) is:
Energy input = ∆H°r / Efficiency
= 205.9 kJ / 0.805
≈ 255.83 kJ

Now let's calculate the energy input for reaction (2): H2O(l) → H2(g) + 1/2 O2(g)

From the given equation, the reaction enthalpy (∆H°r) is +285.8 kJ. This means that 285.8 kJ of energy is required to produce 1 mole of H2 by electrolysis of water. However, this reaction is 49.1% efficient.

Therefore, the energy input for reaction (2) is:
Energy input = ∆H°r / Efficiency
= 285.8 kJ / 0.491
≈ 581.38 kJ

Comparing the energy inputs, we find that reaction (2) requires more energy to produce 1 mole of H2(g) than reaction (1). Hence, the production of H2(g) by electrolysis of water is more energy-intensive compared to the production of H2(g) from methane and water.